A linear object of heigth 10 cm is kept in front of concave mirror of radius of curvature 15 cm, at distance of 10 cm. The image formed is

Options

A. magnified and erect
B. magnified and inverted
C. diminished and erect
D. diminished and inverted

Answer

To solve the problem step by step, we will use the mirror formula and magnification concepts.

### Step 1: Identify the given values
- Height of the object (h₀) = 10 cm
- Radius of curvature (R) = 15 cm
- Object distance (u) = -10 cm (negative because it is in front of the mirror)

### Step 2: Calculate the focal length (f)
The focal length (f) of a concave mirror is given by the formula:
\[ f = \frac{R}{2} \]
Substituting the value of R:
\[ f = \frac{15 \, \text{cm}}{2} = 7.5 \, \text{cm} \]
Since it is a concave mirror, the focal length is negative:
\[ f = -7.5 \, \text{cm} \]

### Step 3: Apply the mirror formula
The mirror formula is given by:
\[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
Substituting the known values:
\[ \frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{-7.5} + \frac{1}{10} \]

### Step 5: Finding a common denominator
The common denominator for -7.5 and 10 is 75. Thus:
\[ \frac{1}{-7.5} = \frac{-10}{75} \]
\[ \frac{1}{10} = \frac{7.5}{75} \]
So,
\[ \frac{1}{v} = \frac{-10 + 7.5}{75} = \frac{-2.5}{75} \]

### Step 6: Solving for v
Now, we can find v:
\[ v = \frac{75}{-2.5} = -30 \, \text{cm} \]

### Step 7: Determine the nature of the image
Since v is negative, the image is formed on the same side as the object, indicating that it is a real image.

### Step 8: Calculate magnification (m)
The magnification (m) is given by:
\[ m = -\frac{v}{u} \]
Substituting the values:
\[ m = -\frac{-30}{-10} = 3 \]
Since the magnification is positive, the image is inverted.

### Step 9: Calculate the height of the image (hᵢ)
Using the magnification formula:
\[ m = \frac{hᵢ}{h₀} \]
Substituting the known values:
\[ 3 = \frac{hᵢ}{10} \]
Thus,
\[ hᵢ = 3 \times 10 = 30 \, \text{cm} \]

### Conclusion
The image formed is real, inverted, and magnified with a height of 30 cm.